The degree three polynomial f(x) with real coefficients and leading coefficient 1, has 4 and 3 + i amoung its?

Question

roots. express f(x) as a product of linear quadratic polynomials with real coefficients.

Answer 1

If a polynome with real coefficients has a complex root a+bi it has also its conjugate a-bi

So your polynome would be P(x) =(x-4)*(x-(3+i))*(x-(3-i))=
=(x-4)*(x^2-6x+10)

Answer 2

first be conscious that one given root is complicated and u might want to understand that complicated and irrational roots continuously happen in conjugate pairs so 3 - i will also be the foundation of this polynomial f(x)=x(x+4)(x^2-(3+i))(x-(3-i)) =x(x+4)(x^2-6x+10) this in linear and quadratic type

Answer 3

The polynomial must be (x-4)(x-3+i)(x-3-i). You do the math.

Answer 4

the "key" is knowing that complex roots ALWAYS come in pairs. So if 3 + i is a root, then you automatically know that 3 - i is also a root.

thus:

f(x) = (x -4)(x -3 +i)(x -3 -i)